"[^] []"是什么意思正则表达式是什么意思? [英] What does the "[^][]" regex mean?

问题描述

我在以下正则表达式中找到了它:

I found it in the following regex:

\[(?:[^][]|(?R))*\]

它将方括号(及其内容)与嵌套方括号匹配.

It matches square brackets (with their content) together with nested square brackets.

推荐答案

[^][]是一个字符类，表示除[和]之外的所有字符.

[^][] is a character class that means all characters except [ and ].

您可以避免转义[和]特殊字符，因为对于preg_函数中使用的正则表达式引擎PCRE而言，这不是模棱两可的.

You can avoid escaping [ and ] special characters since it is not ambiguous for the PCRE, the regex engine used in preg_ functions.

由于[^]在PCRE中不正确，因此正则表达式解析的唯一方法是]在字符类内部，该字符类稍后将关闭.与后面的[相同.它不能重新打开字符类中的字符类(POSIX字符类[:alnum:]除外).然后最后一个]被清除；它是角色类的结尾.但是，字符类外部的[必须转义，因为它被解析为字符类的开头.

Since [^] is incorrect in PCRE, the only way for the regex to parse is that ] is inside the character class which will be closed later. The same with the [ that follows. It can not reopen a character class (except a POSIX character class [:alnum:]) inside a character class. Then the last ] is clear; it is the end of the character class. However, a [ outside a character class must be escaped since it is parsed as the beginning of a character class.

以同样的方式，您可以编写[]]或[[]或[^[]而不在字符类中转义[或].

In the same way, you can write []] or [[] or [^[] without escaping the [ or ] in the character class.

注意:从PHP 7.3开始，您可以使用内联xx修饰符，该修饰符甚至可以在字符类内部忽略空白字符.这样，您可以使用一种不太模糊的方式编写这些类，例如:(?xx) [^ ][ ] [ ] ] [ [ ] [^ [ ].

Note: since PHP 7.3, you can use the inline xx modifier that allows blank characters to be ignored even inside character classes. This way you can write these classes in a less ambigous way like that: (?xx) [^ ][ ] [ ] ] [ [ ] [^ [ ].

您可以将此语法与多种正则表达式一起使用:PCRE(PHP，R)，Perl，Python，Java，.NET，GO，awk，Tcl(如果使用花括号来分隔模式，谢谢Donal Fellows )，...

You can use this syntax with several regex flavour: PCRE (PHP, R), Perl, Python, Java, .NET, GO, awk, Tcl (if you delimit your pattern with curly brackets, thanks Donal Fellows), ...

但不支持:Ruby，JavaScript(除IE< 9 以外的)，...

But not with: Ruby, JavaScript (except for IE < 9), ...

正如m.buettner所指出的，[^]]并不是模棱两可的，因为]是首个字符，[^a]]被视为不是a的所有] .要具有a和]，您必须输入:[^a\]]或[^]a]

As m.buettner noted, [^]] is not ambiguous because ] is the first character, [^a]] is seen as all that is not a a followed by a ]. To have a and ], you must write: [^a\]] or [^]a]

在JavaScript的特殊情况下，规范允许[]作为从不匹配的正则表达式令牌(换句话说，[]总是失败)，而[^]作为匹配的正则表达式任何字符.然后[^]]被视为任何字符，后跟] .实际的实现方式有所不同，但现代浏览器通常会遵循规范中的定义.

In particular case of JavaScript, the specification allow [] as a regex token that never matches (in other words, [] will always fail) and [^] as a regex that matches any character. Then [^]] is seen as any character followed by a ]. The actual implementation varies, but modern browser generally sticks to the definition in the specification.

模式详细信息:

\[          # literal [
(?:         # open a non capturing group
    [^][]   # a character that is not a ] or a [
  |         # OR
    (?R)    # the whole pattern (here is the recursion)
)*          # repeat zero or more time
\]          # a literal ]

在您的模式示例中，您无需转义最后一个]

In your pattern example, you don't need to escape the last ]

但是您可以对此模式进行一些优化，并在子模式(使用(?-1))中重用更有用的原因:(\[(?:[^][]+|(?-1))*+])

But you can do the same with this pattern a little bit optimized, and more useful cause reusable as subpattern (with the (?-1)): (\[(?:[^][]+|(?-1))*+])

(                     # open the capturing group
    \[                # a literal [
        (?:           # open a non-capturing group
            [^][]+    # all characters but ] or [ one or more time
          |           # OR
            (?-1)     # the last opened capturing group (recursion)
                      # (the capture group where you are)
        )*+           # repeat the group zero or more time (possessive)
    ]                 # literal ] (no need to escape)
)                     # close the capturing group

或更好:(\[[^][]*(?:(?-1)[^][]*)*+])避免了更换的费用.

or better: (\[[^][]*(?:(?-1)[^][]*)*+]) that avoids the cost of an alternation.

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